Difference between revisions of "2000 AIME II Problems/Problem 2"

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We must have <math>(x-y)</math> and <math>(x+y)</math> as both even or else x,y would not be an integer. We first give a factor of two to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6*5^6</math> left. Since there are <math>7*7=49</math> factors of <math>2^6*5^6</math>, and since both x and y can be negative, this gives us <math>49\cdot2=98</math> lattice points.
 
We must have <math>(x-y)</math> and <math>(x+y)</math> as both even or else x,y would not be an integer. We first give a factor of two to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6*5^6</math> left. Since there are <math>7*7=49</math> factors of <math>2^6*5^6</math>, and since both x and y can be negative, this gives us <math>49\cdot2=98</math> lattice points.
  
== See also ==
 
 
{{AIME box|year=2000|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2000|n=II|num-b=1|num-a=3}}

Revision as of 20:30, 18 March 2008

Problem

A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$?

Solution

$(x-y)(x+y)=2000^2=2^8*5^6$

We must have $(x-y)$ and $(x+y)$ as both even or else x,y would not be an integer. We first give a factor of two to both $(x-y)$ and $(x+y)$. We have $2^6*5^6$ left. Since there are $7*7=49$ factors of $2^6*5^6$, and since both x and y can be negative, this gives us $49\cdot2=98$ lattice points.

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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